Cho hàm số f(x) thoả mãn lim x suy ra + vô cùng f(x) = 2022. Tính lim x
Giải thích
Ta có \(\mathop {\lim }\limits_{x \to + \infty } \frac{{xf\left( x \right)}}{{x + 1}}\)\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{f\left( x \right)}}{{1 + \frac{1}{x}}} = \frac{{\mathop {\lim }\limits_{x \to + \infty } f\left( x \right)}}{{\mathop {\lim }\limits_{x \to + \infty } \left( {1 + \frac{1}{x}} \right)}}\)
\( = \frac{{\mathop {\lim }\limits_{x \to + \infty } f\left( x \right)}}{{\mathop {\lim }\limits_{x \to + \infty } 1 + \mathop {\lim }\limits_{x \to + \infty } \frac{1}{x}}} = \frac{{2022}}{{1 + 0}} = 2022\).
Vậy \(\mathop {\lim }\limits_{x \to + \infty } \frac{{xf\left( x \right)}}{{x + 1}} = 2022\).